Bit confused on how to complete the question above.
So far I have got this
3tan²θ- 1 = 0
θ = ±√1/3
θ = Pi/6, -Pi/6
Right now this is where I get confused. I was taught to use a this diagram which is shown here ->>Attachment 399589399597
So what do I do now? Because I have looked at the mark scheme which gives two other values of theta. (Solomon paper A)
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- Thread Starter
- 16-05-2015 12:58
- 16-05-2015 13:42
The question asks for an interval where theta is between Pi and -Pi.
Whilst your calculator gives you the result Pi/6 for arctan rt(1/3), there are a lot more solutions for arctan rt(1/3), Pi/6 is the just the most appropriate one. If you look at a tan graph, you can see that it repeats every Pi radians. You can use this to find the other solutions for theta:
(Pi/6) - Pi = (-5Pi/6)
(-Pi/6) + Pi = (5Pi/6)
You can also do (Pi/6) + Pi = (7Pi/6) which would be a solution so long as they were inside the interval, which in this case does not.
I hope you can understand this since I am using the other type of graph that you are used to, but I was never taught to use the type of graph you are using.
- Thread Starter
- 16-05-2015 14:03
This helps but I wont be able to look at a tan graph in my exam. So I wanted to know how to work them out using the diagram attached rather than having to learn the tan graph and sketch it in my exam as this will save me time that I need.Last edited by alexneuty; 16-05-2015 at 14:04.
- 16-05-2015 14:13
From the CAST diagram, you can tell at which angles the value of tan x is the same value you have. As tan is positive in the third quadrant (going anticlockwise) you need to add pi to the principal value you obtained. This gives two values of theta in the interval 0<x<2pi.
For further help, take a look at this: